There are 650 points inside a circle of radius 16. Prove there exists a ring with inner radius 2 and outer radius 3 covering 10 of these points.
Hint of the professor: Use Dirichlet's (pigeonhole) principle.
EDIT: (after the solution became known) Lets call this problem (650, 16, 10). Lets see if we can find another (N, R, M) instances of the problem. (ring (annulus) would stay the same)
According to the solution below, this condition is crucial for proof:
$N\cdot(3^2-2^2)\pi>(M-1)(R+3)^2\pi$ , or
$N>(M-1)(R+3)^2/5$
For $R=16$ and $M=10$, the condition is $N>649.8$.
Lets choose something smaller, $R=5$ and $M=7$, the condition is $N>76.8$.
Or larger, $R=31$ and $M=20$, the condition is $N>4392.8$.
So, following problems are also valid:
There are 77 points inside a circle of radius 5. Prove there exists a ring with inner radius 2 and outer radius 3 covering 7 of these points. (this turned out nice enigmatic combination 7-77-5-2)
There are 4393 points inside a circle of radius 31. Prove there exists a ring with inner radius 2 and outer radius 3 covering 20 of these points. (pretty scary sounding problem)
Around each of the points, draw such an annulus. Then all $650$ annuli are contained within an enlarged circle of radius $19$ and area $19^2\pi$. The sum of the annulus areas is $650\cdot(3^2-2^2)\pi$. This is more than $9\cdot19^2 \pi$ (just slightly: $\frac{650\cdot (3^2-2^2)}{19^2}\approx 9.00277$) , hence some point must be covered at least tenfold. Conversely, the annulus around this point contains at least ten of the original points.